i know this is a kind of big list, but i'm completely lost as to how to do these sort of things and i can't seem to make sense of the book (i've spent about 2 hours trying to figure it out)
any help walking me through these would be greatly appreciated

Prove the following by induction:

A) For all integers n >= 0, the number 5^(2n) - 3n is a multiple of 11

B) Any integer n >= 1, 2^(4n-1) ends with an 8

C) The sum of the cubes of three consecutive positive integers is always a multiple of 9

D) If x >= 2 is a real number and n >= 1 is an integer, then x^n >= nx

E) If n >= 3 is an integer, then 5^n > 4^n + 3^n + 2^n

September 24th 2007, 05:43 PM

ThePerfectHacker

Quote:

Originally Posted by mistykz

i
B) Any integer n >= 1, 2^(4n-1) ends with an 8

Hint: Show that is always divisible by .

September 24th 2007, 06:08 PM

WWTL@WHL

A) Doesn't work. e.g. n=4?

D)

When x=2 and n=1, Therefore true for n=1.
Assume it's true for n=k.

now for induction, we want to prove that
so here is a standard 'trick' often required for similar questions..

For n=k

Times both sides of the inequality by x...

But for x > 0

Add x to both sides.....

Since

It is implied, therefore, that as required.

Hence if n=k is true, then n=k+1 is true. But since n=1 is true, it implies n=2 is true, which implies n=3 is true...etc. Hence proved by induction.

September 24th 2007, 07:26 PM

mistykz

awesome, thanks! anyone else got any others?

September 24th 2007, 07:29 PM

WWTL@WHL

Quote:

Originally Posted by mistykz

awesome, thanks! anyone else got any others?

Can you show your working for any others you've attempted? :)

September 24th 2007, 07:35 PM

mistykz

well, that's the thing. im stuck as to where to start on any of these. he BRIEFLY flew over the stuff, then tossed this homework in our laps. i just sort of need a little coaxing in the right direction

September 25th 2007, 03:50 AM

WWTL@WHL

Well, this is not what you'd call 'elegant', but it's a proof of sorts...
Just use the same 'trick' I did for question D.

E) It's easy to prove that the equation is true for n > 2. So I won't waste time or patronise you by proving that. So assume it is true for n=k. Just deal exclusively with the

multiply each side by 5

But for k > 2

But

(Since )

Therefore it follows that

which implies as required. Hence if n=k is true, then n=k+1 is true. But since n=1 is true, it implies n=2 is true, which implies n=3 is true...etc.

Repeat the process for then . If you want more help, I'd more more than willing to provide it - but can you at least try to say anything you don't understand - or better yet, post some working, that would be great. :)

September 25th 2007, 05:14 AM

Soroban

Hello, mistykz!

Quote:

C) The sum of the cubes of three consecutive positive integers is always a multiple of 9